Question

$\frac{x}{20}M$ concentration of $H^{+}$ ion must be maintained in a saturate $H_{2}S\left(0.1 M\right)$ to precipitate $CdS$ but not $ZnS$ , if $\left[\right.Cd^{2 +} \left]\right.=\left[\right.Zn^{2 +} ⁡ \left]\right.=0.1M$ initially. $K_{s p}\left(\right.CdS \left.\right)=8\times 10^{- 27},K_{s p}\left(Z n S\right)=1\times \left(10\right)^{- 21}K_{a}\left(H_{2} S\right)=1\times \left(10\right)^{- 21}$ . $ZnS$ will not precipitate at a concentration of $H^{+}$ greater than $\frac{x}{20}M$ . The value of $x$ is.

Answer 1

$CdS\rightleftharpoons\underset{0.1}{C d^{2 +}}+\underset{s}{S^{2 -}}$
For precipitation of CdS
$\left[\right.\left(Cd\right)^{2 +} \left]\right.\left[\right.S^{2 -} ⁡ \left]\right.=K_{sp \left(\right. CdS ⁡ \left.\right)} ⁡ $
$0.1\times \left[\right.S^{2 -} \left]\right.=8\times 10^{- 27}$
$\left[\right.S^{2 -} \left]\right.=8\times 10^{- 26}$
For precipitation of ZnS
$\left[\right.\left(Zn\right)^{2 +} \left]\right.\left[\right.S^{2 -} ⁡ \left]\right.=K_{sp \left(\right. ZnS ⁡ \left.\right)} ⁡ $
$0.1\times \left[\right.S^{2 -} \left]\right.=10^{- 21}$
$\left[\right.S^{2 -} \left]\right.=10^{- 20}$
CdS will start precipitate as the conc of $\left[\right.S^{2 -} \left]\right.=8\times 10^{- 26}$ and ZnS will start precipitate as the conc of $\left[\right.S^{2 -} \left]\right.=10^{- 20}$ so we can raise the concentration of $\left[\right.S^{2 -} \left]\right.$ upto $10^{- 20}$
Now
$\underset{\underset{\approx 0.1}{0.1 \left(\right. 1 - \alpha \left.\right)}}{H_{2} S}\rightleftharpoons\underset{\underset{= \frac{x}{20} M}{2 \times 0.1 \alpha }}{2 H^{+}}+\underset{0.1 \alpha = 1 0^{- 20}}{S^{2 -}}$
$\frac{\left(\frac{x}{20}\right)^{2} \times 1 0^{- 20}}{0.1}=10^{- 21}$
$\left(\frac{x}{20}\right)^{2}=10^{- 2}$
$\frac{x}{20}=10^{- 1}$
$x=2$